Discrete Mathematics: Elementary and Beyond L. Lovász J. Pelikán K. Vesztergombi Springer [...]... Subsets 11 contains the third (and in this case last) question you have to answer to determine the subset: Is c an element of S? Giving an answer and following the appropriate arrow we finally get to a node that does not represent a question, but contains a listing of the elements of S Thus to select a subset corresponds to walking down this diagram from the top to the bottom There are just as many subsets... Frank “This means 2 · 2 · 2 = 8 ways to rearrange people without changing the pairing So in fact, there are 6 · 8 = 48 ways to sit that all mean the same pairing The 720 seatings come in groups of 48, and so the number of matchings is 720/48 = 15.” “I think there is another way to get this,” says Alice after a little time “Bob is youngest, so let him choose a partner first He can choose his partner in. .. , an Then we may or may not want to include a1 , in other words, we can make two possible decisions at this point No matter how we decided about a1 , we may or may not want to include a2 in the subset; this means two possible decisions, and so the number of ways we can decide about a1 and a2 is 2 · 2 = 4 Now no matter how we decide about a1 and a2 , we have to decide about a3 , and we can again decide... Alice invites six guests to her birthday party: Bob, Carl, Diane, Eve, Frank, and George When they arrive, they shake hands with each other (strange European custom) This group is strange anyway, because one of them asks, “How many handshakes does this mean?” “I shook 6 hands altogether,” says Bob, “and I guess, so did everybody else.” “Since there are seven of us, this should mean 7 · 6 = 42 handshakes,”... the room, and so they never saw a real tree.) We can give another proof of Theorem 1.3.1 Again, the answer will be made clear by asking a question about subsets But now we don’t want to select a subset; what we want is to enumerate subsets, which means that we want to label them with numbers 0, 1, 2, so that we can speak, say, about subset number 23 of the set In other words, we want to arrange the... left to the reader as an exercise Warning! Before going too far with this analogy, let us point out that there is another distributive law for sets: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (1.4) 8 1 Let’s Count! We get this simply by interchanging “union” and “intersection” in (1.1) (This identity can be proved just like (1.1); see Exercise 1.2.16.) This second distributivity is something that has no analogue... they had to decide (remember, this happens in a poor European country) that they didn’t have enough money to buy so many tickets (Besides, they would win much less And to fill out so many tickets would spoil the party!) So they decide to play cards instead Alice, Bob, Carl and Diane play bridge Looking at his cards, Carl says, “I think I had the same hand last time.” “That is very unlikely” says Diane How... proofs are too long or too involved to be included in this introductory book So why did we bother to give any proof, let alone two proofs of the same statement? The answer is that every proof reveals much more than just the bare fact stated in the theorem, and this revelation may be more valuable than the theorem itself For example, the first proof given above introduced the idea of breaking down the... to denote logarithm with base 10) We have then k − 1 ≤ x < k, which means that k − 1 is the largest integer not exceeding x Mathematicians have a name for this: It is the integer part, or floor, of x, and it is denoted by x We can also say that we obtain k by rounding x down to the next integer There is also a name for the number obtained by rounding x up to the next integer: It is called the ceiling... discussed during the party were special cases of theorem 1.8.1? 1.8.2 Tabulate the values of n k for 0 ≤ k ≤ n ≤ 5 1.8.3 Find the values of n for k = 0, 1, n − 1, n using (1.6), and explain the k results in terms of the combinatorial meaning of n k Binomial coefficients satisfy many important identities In the next theorem we collect some of these; some other identities will occur in the exercises and in the . adds Frank. “This means 2 · 2 · 2 = 8 ways to rearrange people without changing the pairing. So in fact, there are 6 · 8 = 48 ways to sit that all mean the same pairing. The 720 seatings come in. shake hands with each other (strange European custom). This group is strange anyway, because one of them asks, “How many handshakes does this mean?” “I shook 6 hands altogether,” says Bob, “and. 153 9 Finding the Optimum 157 9.1 Finding the Best Tree 157 9.2 The Traveling Salesman Problem 161 10 Matchings in Graphs 165 Contents ix 10.1 A Dancing Problem 165 10.2 Another matching problem